The car’s velocity at the end of this distance is <em>18.17 m/s.</em>
Given the following data:
- Initial velocity, U = 22 m/s
- Deceleration, d = 1.4
To find the car’s velocity at the end of this distance, we would use the third equation of motion;
Mathematically, the third equation of motion is calculated by using the formula;
Substituting the values into the formula, we have;
<em>Final velocity, V = 18.17 m/s</em>
Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>
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Read more: brainly.com/question/8898885
Answer:
Value that the spring constant k = 12Mg / h
Explanation:
According to 2nd law of Newton:
upward force of the spring= F
The weight of the elevator W = mg
F = Mg = M(5g)
==> F =6Mg.
As the spring is compressed to its maximum distance ie s,the maximum upward acceleration comes just , Hence
F =ks = 6Mg
==> s = 6Mg/k
We have gravitational potential energy turning into elastic potential of the spring as the elevator starts at the top some distance h from the spring, and undergoes a total change in height equal to h + s, so:
Mg(h+s) = 1/2ks2
And plugging in our expression for s:
Mg(h+6Mg/k)= 1/2k(6Mg / k)2
gh + 6M2g2/k = 1/2k(36M2g2 /k2)
Mgh +6M2g2/k = 1/2k(36M2g2 /k2)
gh + 6Mg2/k = 18Mg2 / k
gh = 12Mg2 / k
h = 12Mg / k
k = 12Mg / h
charge stored in the capacitor=3.29 x 10⁻⁴ C
Explanation:
we use the formula
Q= C V
Q= charge
C= capacitor=25.3 μF= 25.3 x 10⁻⁶ F
V= voltage= 13 V
Q=(25.3 x 10⁻⁶ ) (13)
Q= 3.29 x 10⁻⁴ C
Your increasing the volume on mp3 player
final velocity = 0
acceleration = - 10 m/ s 2
distance. = 20 m
u = ?
v^2 - u ^2 = 2 a s
0^2 - u^ 2 = 2 * -10 * 20
-u^2 = -400
u = √ 400
u = 20m / s
